YES Termination w.r.t. Q proof of TCT_12_recursion-5.ari

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f_0(x) → a
f_1(x) → g_1(x, x)
g_1(s(x), y) → b(f_0(y), g_1(x, y))
f_2(x) → g_2(x, x)
g_2(s(x), y) → b(f_1(y), g_2(x, y))
f_3(x) → g_3(x, x)
g_3(s(x), y) → b(f_2(y), g_3(x, y))
f_4(x) → g_4(x, x)
g_4(s(x), y) → b(f_3(y), g_4(x, y))
f_5(x) → g_5(x, x)
g_5(s(x), y) → b(f_4(y), g_5(x, y))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:
f51 > g52 > f41 > [s1, g42] > f31 > g32 > f21 > [f11, g22] > g12 > [f01, a, b2]

Status:
f01: multiset
a: multiset
f11: multiset
g12: [1,2]
s1: [1]
b2: [1,2]
f21: multiset
g22: multiset
f31: multiset
g32: [1,2]
f41: multiset
g42: [1,2]
f51: multiset
g52: [1,2]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f_0(x) → a
f_1(x) → g_1(x, x)
g_1(s(x), y) → b(f_0(y), g_1(x, y))
f_2(x) → g_2(x, x)
g_2(s(x), y) → b(f_1(y), g_2(x, y))
f_3(x) → g_3(x, x)
g_3(s(x), y) → b(f_2(y), g_3(x, y))
f_4(x) → g_4(x, x)
g_4(s(x), y) → b(f_3(y), g_4(x, y))
f_5(x) → g_5(x, x)
g_5(s(x), y) → b(f_4(y), g_5(x, y))


(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) YES